Normal is perpendicular to tangent at the pint on the curve.

The slope of tangent to curve #y=f(x)# at #x=x+# is given by #f'(x_0)# and hence slope of normal is #-1/(f'(x_0))#. Further itpasses through point #(x_0,f(x_0))# on the curve. This gives us point slope form of normal to the curve as #y-f(x_0)=-1/(f'(x_0))(x-x_0)#.

Here #y=f(x)=sinx-e^xcotx# and #x_0=pi/3#

Hence #f(pi/3)=sin(pi/3)-e^(pi/3)cot(pi/3)=sqrt3/2-e^(pi/3)/sqrt3#

Further #f'(x)=cosx+e^xcsc^2x-e^xcotx#

and #f'(pi/3)=cos(pi/3)+e^(pi/3)csc^2(pi/3)-e^(pi/3)cot(pi/3)#

= #1/2+e^(pi/3)[4/3-1/sqrt3]=1/2+(4-sqrt3)/3e^(pi/3)#

and hence equation of normal is

#y-sqrt3/2+e^(pi/3)/sqrt3=-1/(1/2+(4-sqrt3)/3e^(pi/3))(x-pi/3)#

or #(y-sqrt3/2+e^(pi/3)/sqrt3)(1/2+(4-sqrt3)/3e^(pi/3))=pi/3-x#